AIME II #10 2003

[stupidityismygam]

Two positive integers differ by [unparseable or potentially dangerous latex formula]. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

[mathslug]

This problem was annoying, and I need someone to crosscheck me, but I think I got it. let N be the non-perfect square.

a-b=60 therefore, a = 60+b

(sqrt(60+b)+sqrt(b)) = sqrt(N) square and reduce until you get ===>

0 = N^2 - (4b+120)N + 3600

Let x and y be the roots for N.

xy = 3600 and x+y = 4b+120 rearrange this second eq to get (x+y)/4 -30 = b, which means that the sum of the roots (and in this case, number pairs that divide into 3600) must be divisible by 4 to make b an INTEGER. We want N to be an integer as well, so lets go through the first few pairs:

1 - 3600
2 - 1800
3 - 1200
4 - 900
5 - 720
6 - 600
8 - 450
9 - 400
10 - 360
12 - 300
15 - 240
16 - 225
18 - 200
20 - 180
24 - 150
25 - 144
30 - 120
36 - 100
40 - 90
45 - 80
48 - 75
50 - 72
60 - 60

The biggest b is the one we want to answer this question, which should be the first sum that is divisible by 4 that is NOT a perfect square, because it said not to use those. The first pair on this list that rings that bell is 300 and 12.

b = 312/4 -30 = 48, and a = 108

sqrt(108) + sqrt(48) = sqrt(300)

CHECK and MATE

The largest possible sum, therefore, is 156

Zack

[stupidityismygam]

yep yep yep good job sir